Theorem freq2 4377

Description: Equality theorem for the well-founded predicate. (Contributed by NM, 3-Apr-1994.)

Assertion

Ref	Expression
freq2	⊢ (𝐴 = 𝐵 → (𝑅 Fr 𝐴 ↔ 𝑅 Fr 𝐵))

Proof of Theorem freq2

Dummy variable 𝑠 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		frforeq2 4376	. . 3 ⊢ (𝐴 = 𝐵 → ( FrFor 𝑅𝐴𝑠 ↔ FrFor 𝑅𝐵𝑠))
2	1	albidv 1835	. 2 ⊢ (𝐴 = 𝐵 → (∀𝑠 FrFor 𝑅𝐴𝑠 ↔ ∀𝑠 FrFor 𝑅𝐵𝑠))
3		df-frind 4363	. 2 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑠 FrFor 𝑅𝐴𝑠)
4		df-frind 4363	. 2 ⊢ (𝑅 Fr 𝐵 ↔ ∀𝑠 FrFor 𝑅𝐵𝑠)
5	2, 3, 4	3bitr4g 223	1 ⊢ (𝐴 = 𝐵 → (𝑅 Fr 𝐴 ↔ 𝑅 Fr 𝐵))

Syntax hints:   → wi 4   ↔ wb 105  ∀wal 1362   = wceq 1364   FrFor wfrfor 4358   Fr wfr 4359

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-ral 2477  df-in 3159  df-ss 3166  df-frfor 4362  df-frind 4363

This theorem is referenced by:  weeq2  4388